If `tan^(2) {pi(x+y)}+cot^(2) {pi (x+y)}=1+sqrt((2x)/(1+x^(2)))` where `x, y in R`, then find the least possible value of y.

Explanation

To solve the equation \( \tan^2(\pi(x+y)) + \cot^2(\pi(x+y)) = 1 + \sqrt{\frac{2x}{1+x^2}} \) and find the least possible value of \( y \), we can follow these steps:

### Step 1: Analyze the Left-Hand Side (LHS)

The left-hand side of the equation is \( \tan^2(\pi(x+y)) + \cot^2(\pi(x+y)) \). We know that:

\[ \tan^2 a + \cot^2 a \geq 2 \]

This inequality holds true for any angle \( a \), and the equality occurs when \( a = \frac{\pi}{4} \) (or \( \tan a = 1 \)). Therefore, we can conclude:

\[ \tan^2(\pi(x+y)) + \cot^2(\pi(x+y)) \geq 2 \]

### Step 2: Analyze the Right-Hand Side (RHS)

Now, let's analyze the right-hand side:

\[ 1 + \sqrt{\frac{2x}{1+x^2}} \]

To find the maximum value of \( \sqrt{\frac{2x}{1+x^2}} \), we can set \( f(x) = \frac{2x}{1+x^2} \) and find its maximum.

### Step 3: Find the Maximum of \( \frac{2x}{1+x^2} \)

To find the maximum value of \( f(x) \), we can differentiate it:

\[ f'(x) = \frac{(1+x^2)(2) - 2x(2x)}{(1+x^2)^2} = \frac{2(1+x^2 - 2x^2)}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \]

Setting \( f'(x) = 0 \):

\[ 1 - x^2 = 0 \implies x = 1 \text{ (since } x \in \mathbb{R}\text{)} \]

Now, we evaluate \( f(1) \):

\[ f(1) = \frac{2 \cdot 1}{1 + 1^2} = \frac{2}{2} = 1 \]

Thus, the maximum value of \( \sqrt{\frac{2x}{1+x^2}} \) is \( 1 \) when \( x = 1 \).

### Step 4: Substitute Back into the RHS

Substituting this back into the RHS:

\[ 1 + \sqrt{\frac{2x}{1+x^2}} \leq 1 + 1 = 2 \]

### Step 5: Set LHS Equal to RHS

Since both sides can equal \( 2 \), we set:

\[ \tan^2(\pi(x+y)) + \cot^2(\pi(x+y)) = 2 \]

This equality holds when \( \tan(\pi(x+y)) = 1 \), which occurs when:

\[ \pi(x+y) = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 6: Solve for \( y \)

From \( \pi(x+y) = \frac{\pi}{4} + n\pi \):

\[ x + y = \frac{1}{4} + n \]

Thus,

\[ y = \frac{1}{4} + n - x \]

### Step 7: Find the Least Possible Value of \( y \)

To minimize \( y \), we can choose \( n = 0 \) and \( x = 1 \):

\[ y = \frac{1}{4} + 0 - 1 = \frac{1}{4} - 1 = -\frac{3}{4} \]

### Conclusion

The least possible value of \( y \) is:

\[ \boxed{-\frac{3}{4}} \]